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Chapter 7: Problem 61

Find the volume of the following solids. The region bounded by \(y=1 /(x+2), y=0, x=0,\) and \(x=3\) is revolved about the line \(x=-1\)

Short Answer

Expert verified
Question: Determine the volume of the solid formed when the region bounded by the curves \(y = \frac{1}{x + 2}\), \(y = 0\), \(x = 0\), and \(x = 3\) is revolved around the line \(x = -1\). Answer: The volume of the solid is \(V = 21\pi\).

Step by step solution

01

Understanding the Disc Method

When a region in the xy-plane is revolved around a line, it forms a solid with circular cross-sections perpendicular to the line it is being revolved around. The disc method can be used to find the volume of this solid by integrating the areas of these circular cross-sections along the axis of revolution. In this case, the region is bounded by \(y=1/(x+2), y=0, x=0,\) and \(x=3\), and it is revolved around the line \(x=-1\). Since each cross-section is a disc, we can find the area of each disc and then integrate over the range of x values in the given region.
02

Determine the Range for Integration

As the region is defined by the curves \(y=1/(x+2), y=0, x=0,\) and \(x=3\), we will integrate over the x values. The limits of integration for x will be from x = 0 to x = 3.
03

Find the Radius Function

To use the disc method, we need to find the radius function in terms of x. The radius of each disc will be the distance between the curve \(y=1/(x+2)\) and the line \(x=-1\). Using the distance formula: \(r(x) = (x - (-1)) = (x + 1)\) Now we can find the area of each disc as a function of x: \(A(x) = \pi [r(x)]^2 = \pi (x + 1)^2\)
04

Set Up the Integral

To find the volume of the solid, we integrate the area function over the range of x: \(V = \int_{x=0}^{x=3} A(x) \,dx = \int_{0}^{3} \pi(x+1)^2 \,dx\)
05

Evaluate the Integral

To evaluate the integral, we first need to expand the integrand: \((x + 1)^2 = x^2 + 2x + 1\) Now, integrate term by term: \(\int_{0}^{3} \pi(x^2 + 2x + 1) \,dx = \pi \int_{0}^{3} (x^2 + 2x + 1) \,dx\) Using the power rule and the rules for integrating polynomials: \(= \pi [(\frac{1}{3}x^3 + x^2 + x)|_0^3]\) Evaluate the integral at the limits: \(= \pi [(\frac{1}{3}(3^3) + (3^2) + 3) - (\frac{1}{3}(0^3) + (0^2) + 0)]\) \(= \pi [(\frac{1}{3}(27) + 9 + 3)]\) \(= \pi [9 + 9 + 3]\) \(= \pi [21]\) So the volume of the solid is: \(V = 21\pi\)

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